9
$\begingroup$

The du Val singularities are the simplest type of surface singularities. Each type of du Val singularity has a divisor class group. Specifically, let $X$ be a surface with an isolated singularity at $P$; then the (analytic or étale) local ring at P depends only on the type of the singularity, and has a divisor class group.

The most familiar example is the quadric cone (A1 singularity), found in many algebraic geometry textbooks. A line $L$ passing though the vertex of the cone is not locally principal, but $2L$ is, and we find that the divisor class group has order $2$. (Note: in general an A1 singularity will be étale locally, but not Zariski locally, isomorphic to the vertex of the cone. As far as I can see, there's no reason in general to expect the generator of the divisor class group to come from a divisor on the ambient surface; we may well have to pass to an étale (or analytic) neighbourhood.)

In a beautiful article, Lipman (Pub. Math. IHES 1969) studied these and computed the (finite) divisor class group of each du Val singularity. However, as far as I can see, he does not give explicit generators like we have in the example of the quadric cone.

So:

Is there in the literature an explicit description (i.e. with explicit generators) of the divisor class groups of the du Val singularities?

$\endgroup$

2 Answers 2

5
$\begingroup$

You can mimic the quadric cone construction (if I did not make any mistakes in my computation). An $A_{2k-1}$ singularity is the vertex of the cone $S$ given by $x^2+y^2+z^{2k}=0$ in the weighted projective space $P(k,k,1,1)$ (note that the ambient wps is smooth at the vertex of $S$). Any point $p$ on the curve $C$ given by $x^2+y^2+z^{2k}=0$ in $P(k,k,1)$ yields a rational curve on $S$, which is not a principal divisor. Is this enough for your purposes?

You can do something similar for arbitrary $A_{2k}$, $D_m$ and $E_n$ singularities.

$\endgroup$
1
  • $\begingroup$ Thanks, Remke - I hadn't considered weighted projective space. I'll have a play and see whether this works. $\endgroup$ Aug 11, 2011 at 22:13
4
$\begingroup$

I do not know a reference, but here is my guess. I will use the notion of the wikipedia article. The order is: type, class group, the generator ideals. ($i^2=-1$, and I assume char. 0 for simplicity)

$A_n$, $\mathbb Z/(n+1)$, $(w+ix, y)$.

$D_n$ ($n$ even), $\mathbb Z/(2)\oplus \mathbb Z/(2)$, $(w,y), (w, x+iy^{(n-2)/2})$.

$D_n$ ($n$ odd), $\mathbb Z/(4)$, $(w,y)$.

$E_6$, $\mathbb Z/(3)$, $(x, w+iy^2)$.

$E_7$, $\mathbb Z/(2)$, $(w,x)$.

$E_8$, $0$.

$\endgroup$
1
  • $\begingroup$ Thank you. I find it interesting that you don't have to pass to an étale cover in any of these cases. $\endgroup$ Aug 11, 2011 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.